How do you write 4540 million in standard form?

2 answers:

8 0

For this case what we should do is take into account the following conversion:
$million = 10 ^ 6
$
We then have the following number:
$4540 million
$
Applying the given conversion we have:
$4540 * 10 ^ 6
$
By doing the corresponding calculation we have that the standard form is given by:
$4540 * 10 ^ 6 = 4,540,000,000
$
Answer:
4540 million in standard form is:
$4,540,000,000$

Shalnov [3]2 years ago

7 0

$4540mln=4,540,000,000\\\\in\ scientific\ notation:4,\underbrace{540,000,000}_{\leftarrow9}=4.54\cdot10^9$

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A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defe

xxMikexx [17]

Answer:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

Step-by-step explanation:

For the random variable $N_1$ we define the possible values for this variable on this case $[1,2,3,4,5]$ . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

$5C2 = \frac{5!}{2! (5-2)!}= \frac{5*4*3!}{2! 3!}= \frac{5*4}{2*1}=10$

We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable $N_1$ like this:

$P(N_1 = a) = \frac{5-a C 1}{5C2}$

For the distribution of $N_2$ we need to take in count that we are finding a conditional distribution. $N_2$ given $N_1 =a$ , for this case we see that $N_2 \in [1,2,...,5-a]$ , so then exist $5-a C 1$ ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined: